How do you factor #16x^2 - 8x - 15#?

1 Answer
Jan 10, 2016

Answer:

y = (4x + 3)(4x - 5)

Explanation:

I use the New AC Method to factor trinomials (Socratic Search)
#y = 16x^2 - 8x - 15 = #16(x + p)(x + q) (1)
Converted trinomial: #y' = x^2 - 8x - 240 =# (x + p')(x + q') (2).
p' and q' have opposite signs.
Factor pairs of (ac = - 240) --> (-6, 40)(-8, 30)(-12, 20). This sum is: (20 - 12 = 8 = -b). The opposite sum (12, -20) gives: p' = 12 and q' = -20. Back to trinomial (1), #p = (p')/a = 12/16 = 3/4# and #q = (q')/a = -20/16 = -5/4.#
Factored form: #y = 16(x + 3/4)(x - 5/4) = (4x + 3)(4x - 5)#.
Check by development:
#(4x + 3)(4x - 5) = 16x^2 - 20x + 12x - 15 = 16x^2 - 8x - 15#. OK